Math Is Fun Forum

ninjaman

Member

Registered: 2013-10-15

Posts: 61

evaluate integration

hello

s = integration sign (top number, bottom number)

i have this :

s (1, 0) (2xe^3x dx)

u = 2x

du/dx = x/2

dv/dx = e^3x

v = s 1/3 ^e3x

i really have no clue what im doing.
if someone could tell me what i do any of this it would really help

thanks
simon:)

i have 2x so to integrate it, i increase the power by 1 and divide by this figure. so 2x = 2x^1, add 1 = 2x^2/2 (correct?)

is 2xe^3x treated as one number, like 2*xe^3x. one answer the lecturer wrote was 2e^3x/3, = 2/3 s e^3x dx

i am looking on the internet at the moment for any answers and coming up with nothing.
does anyone know any good sites that explain with good, easy to follow examples.

thanks

Last edited by ninjaman (2014-03-03 01:05:04)

Bob

Administrator

Registered: 2010-06-20

Posts: 10,143

Re: evaluate integration

hi Simon

When you have to integrate two functions multiplied like this, you can often use 'integration by parts'.  You'll find lots of sites with help on this.  It comes from the product rule in differentiation.

The functions are 2x and e^(3x).  Look to see if one will get easier when differentiated and the other no more complicated when integrated.

In this case they are in the right order for exactly that so put u = 2x and dv/dx = e^(3x)

Then you change the integration like this:

so if u = 2x, then du/dx = 2

and dv/dx = e^(3x) then v = 1/3 of e^(3x)

Substitute in the formula and your integral becomes

This second part is fairly easy to integrate and then you apply the limits to both parts.

http://en.wikipedia.org/wiki/Integration_by_parts

http://www.khanacademy.org/math/calculu … ts-formula

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

ninjaman

Member

Registered: 2013-10-15

Posts: 61

Re: evaluate integration

ive tried to add to this, i have read through what i have found and im still confused

Bob

Administrator

Registered: 2010-06-20

Posts: 10,143

Re: evaluate integration

hi ninjaman,

Let's say you have a function y that is made by multiplying two simpler functions of x together. 

If I call those functions u and v then, using the product rule:

Therefore

This makes a formula that can be used to convert one integral into another

It can sometimes be used when you have two functions of x multiplied together.  You call one of them 'v' and the other du/dx.

To use the formula you also have to work out 'u' and dv/dx, so it is only worth doing when u is easy to work out and dv/dx is something simpler.  That can make a hard integral into an easier one.

eg. 

Looking at this I can see that the integral of sin(x) is just another trig function, so that doesn't make the problem any worse, and x is easy to differentiate.  So I'll choose

v = x    and    du/dx = sin(x)

Now work out dv/dx = 1     and u = -cos(x)

Now I can substitute into the formula:

becomes

If there are limits you substitute them as usual.  Otherwise this would need a constant of integration as well.

check.  Any method of integration is OK if it works.  To test this, take the answer and differentiate it.  Do you get the original function?

That's what I should have got, so the integral is correct.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

ninjaman

Member

Registered: 2013-10-15

Posts: 61

Re: evaluate integration

hello

I ended up with loge 1 * 1/3 e ^3*1 and, loge 0 * 1/3 e ^3*0
I also put those upper and lower limits in the x in the [] 2x/3 = 2*1/3 and 2*0/3
and e^3*1, e^3*0
any help

im fed up of this

thanks
simon:)

Last edited by ninjaman (2014-03-20 10:58:38)

eigenguy

Member

Registered: 2014-03-18

Posts: 78

Re: evaluate integration

The problem is to evaluate

As bob as already indicated

Hopefully, you should now (bob already used this fact once) that

So putting it together gives:

Now put in your limits of integration (where I can drop the "+ C", as it cancels out):

It still needs a bit more simplification, but hopefully this will straighten out where you went wrong.

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"Having thus refreshed ourselves in the oasis of a proof, we now turn again into the desert of definitions." - Bröcker & Jänich